what happened on march 21 every year in the northern hemisphere​

What Happened On March 21 Every Year In The Northern Hemisphere

Answers

Answer 1

Answer:

B. The Spring equinox

Explanation:

The vernal equinox marks the moment the sun crosses the celestial equator.  The vernal equinox happens on March 19, 20, or 21 every year in the Northern Hemisphere. In the Southern Hemisphere, this same event marks the beginning of fall.  (Source: What Exactly Is The Spring Equinox? - Dictionary.com)

Hopefully this helps.


Related Questions

At a rock concert, the sound intensity 1.0m in front of the bank of loudspeakers is 0.10 W/m^2. A fan is 30.0m from the loudspeakers. Her eardrums have a diameter of 8.4mm.How much energy is transferred to each eardrum in one second?

Answers

Answer: The amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].

Explanation:

Given: Intensity = 0.10 [tex]W/m^{2}[/tex]

Now, at 30.0 m the intensity will be calculated as follows.

[tex]Intensity = \frac{(30.0 m)^{2}}{(1 m)^{2}}\\= 900[/tex]

This means that the intensity is 900 times less assuming that the wave is spherical.

Hence, the new intensity is calculated as follows.

[tex]\frac{0.10 W/m^{2}}{900}\\= 1.11 \times 10^{-4} W/m^{2}[/tex]

The area of ear drum is expressed as follows.

[tex]Area = \pi \times r^{2}\\= 3.14 \times (4.2 mm)^{2}\\= 5.54 \times 10^{-5} m^{2}\\[/tex]

Formula used to calculate energy is as follows.

[tex]P = I \times a[/tex]

where,

P = power or energy

I = intensity

a = area

Substitute the values into above formula as follows.

[tex]P = I \times a\\= 1.11 \times 10^{-4} W/m^{2} \times 5.54 \times 10^{-5} m^{2}\\= 6.2 \times 10^{-9} W[/tex]

Thus, we can conclude that amount of energy transferred to each eardrum in one second is [tex]6.2 \times 10^{-9} J[/tex].

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval

Answers

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

Computer use ___code to transmit information

Answers

Binary code is the answer

Answer:

binary code is the answer of blank

URGENT

The component of solid-state physics that works with and/or heats metals and alloys to give them certain desired
shapes or properties is..

Answers

Answer:

the is metallurgy .....

A hockey puck is sliding across the ice with an initial velocity of 25 m/s. If the coefficient of friction between the hockey puck and the ice is 0.08, how much time (in seconds) will it take before the hockey puck slides to a stop

Answers

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.

Answers

Answer:

a) S =  63.2 km/h

b) V =  63.2 km/h*(-0.316 , 0.949)

Explanation:

Let's define:

North as the positive y-axis

East as the positive x-axis.

Also, remember the relation:

Distance = Time*Speed

Let's assume that she starts at the position (0km, 0km)

Then she travels due North at 90km/h for two hours, then the displacement is

90km/h*2h = 180km to the north

Then the new position is:

(0km, 180km)

Then she travels West at 60km/h for one hour.

Then the distance traveled to the West (negative x-axis) is:

60km/h*1h = 60km to the west

Then the new position is:

(-60km, 180km).

a) The average speed is defined as the quotient between the displacement and the time.

We know that the total time traveled is 3 hours.

And the displacement is the difference between the final position and the initial position.

this is:

D = √( -60km - 0km)^2 + (180km - 0km)^2)=

D = √( (60km)^2 + (180km)^2) = 189.7 km

Then the average speed is:

S = (189.7 km)/(3 h) = 63.2 km/h

b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.

So, if the final position is (-60km, 180km)

We need to find a vector that represents the same angle, but that is on the unit circle.

Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:

(-60km/ 189.7 km, 180km/ 189.7 km)

(-60/189.7 , 180/189.7)

We can just check that the module of the above versor is 1.

[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]

Then the average velocity is:

V = 63.2 km/h*(-60/189.7 , 180/189.7)

We can simplify our versor so the velocity equation is easier to read:

V = 63.2 km/h*(-0.316 , 0.949)

You want to calculate how long it takes a ball to fall to the ground from a
height of 20 m. Which equation can you use to calculate the time? (Assume
no air resistance.)
O A. vz? = v? +2aAd
B. a =
V₂-vi
At
O c. At=V1
4
a
O D. At=
2Ad
a

Answers

If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:

D. Δt= √2Δd/a

What is the equation for finding the length of time for a free fall?

The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².

So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.

Learn more about free fall here:

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The liquid and gaseous state of hydrogen are in thermal equilibrium at 20.3 K. Even though it is on the point of condensation, model the gas as ideal and determine the most probable speed of the molecules (in m/s). What If? At what temperature (in K) would an atom of xenon in a canister of xenon gas have the same most probable speed as the hydrogen in thermal equilibrium at 20.3 K?

Answers

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

Example 9.1
The Archer
Let us consider the situation proposed at the beginning of
this section. 160kg archer stands at rest on frictionless ice
and fires a 0.50-kg arrow horizontally at 50 m s (Fig. 9.2).
With what velocity does the archer move across the ice after
firing the arrow​

Answers

v1f = -0.16 ms

Explanation:

Use the conservation law of linear momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where

v1i = v2i = 0

m1 = 160 kg

m2 = 0.50 kg

v2f = 50m/s

v1f = ?

So we have

0 = (160 kg)v1f + (0.5 kg)(50 m/s)

v1f = -(25 kg-m/s)/(160 kg)

= -0.16 m/s

Note: the negative sign means that its direction is opposite that of the arrow.

Can anyone help
Me please the question is on the photo that I attached it to

Answers

Answer:

2.8 MW

Explanation:

There are 7 wind turbines in the wind farm as shown in the diagram. Thus, the energy output by one turbine is 1/7 if the total energy output. So, 19.6/7=2.8MW

Consider a sample containing 1.70 mol of an ideal diatomic gas.
(a) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(b) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
(c) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(d) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K

Answers

I don't know

because I don't know

3. Four charges having charge q are placed at the corners of a square with sides of length L. What is the magnitude of the force acting on any of the charges

Answers

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.

Answers

Answer:

Explanation:

Given that:

Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm

For objective lens of focal length f₁ = 20 mm

s₁' = 120 mm + 20 mm = 140 mm

Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{140}{20}[/tex]

[tex]m_1 = 7 \ m[/tex]

For objective lens of focal length f₁ = 4 mm

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{124}{4}[/tex]

[tex]m_1 = 31 \ m[/tex]

For objective lens of focal length f₁ = 1.4 mm

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{121.4}{1.4}[/tex]

[tex]m_1 = 86.71 \ m[/tex]

The magnification of the eyepiece is given as:

[tex]m_e = 5X \ and \ m_e = 15X[/tex]

Thus, the largest angular magnification when  [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]

[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]

= 86.71 × 15

= 1300.65

The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]

[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]

= 7 × 5

= 35

The largest magnification will be 1300.65 and the smallest magnification will be 35.

What is magnification?

Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.

Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.

It is given in the question:

Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm

now we will calculate the magnification for all three focal lengths of the objective lens.

Also, each objective forms an image 120 mm beyond its second focal point.

(1) For an objective lens of focal length   [tex]f_1=20 \ mm[/tex]

[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]

Magnification will be calculated as

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]

(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]

(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]

Now the magnification of the eyepiece is given as:

[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]

Thus, the largest angular magnification when  

[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]

[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]

[tex]m_{large}=86.71\times 15=1300.65[/tex]

The smallest angular magnification derived when

[tex]m_1=7\ \ \ \ m_e=5[/tex]

[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]

[tex]m_{small}=7\times 5=35[/tex]

Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.

To know more about magnification follow

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Mention & Instrument used to measure
the mass of the body.

Answers

Answer:

a scale is used to measure the mass of the body

who is corazon aquino? ​

Answers

Answer:

Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.

Answer:

Former President of the Philippines

Explanation:

En 2.0 s, una particula con aceleración constante a lo largo del eje x se mueve desde x =10 m
hasta x =50 m. La rapidez al final del recorrido es de 10 m/s. ¿Cuál es la aceleración de la partícula?

Answers

The correct response is x23

What are the practical systems under continuos time system?​

Answers

Answer:

REVIEWS

Wing-Kuen Ling, in Nonlinear Digital Filters, 2007

Controllability and observability

For a continuous time system, assume that x(0) = 0. ∀x1, if ∃t1 > 0 and u(t) such that x(t1) = x1, then the continuous time system is said to be reachable. Similarly, for a discrete time system, assume that x(0) = 0. ∀x1, if ∃n1 > 0 and u(n) such that x(n1) = x1, then the discrete time system is said to be reachable. For a continuous time system, if ∀x0, x1, ∃t1 > 0 and u(t) such that x(0) = x0 and x(t1) = x1, then the continuous time system is said to be controllable. Similarly, for a discrete time system, if ∀x0, x1, ∃n1 > 0 and u(n) such that x(0) = x0 and x(n1 = x1, then the discrete time system is said to be controllable. For LTI systems, the set of reachable state is R(|BAB… AnB|), where R(A) is defined as the range of A, that is R(A) = {y : y = Ax}. Also, the LTI systems are controllable if and only if R(A) = Rn Or in other words, rank(|BAB… AnB|) = n.

How can i prove the conservation of mechanical energy?​

Answers

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass​

Answers

Answer:

The product of mass and velocity is the correct answer

Explanation:

Momentum is defined as mass × velocity

p = mv

Answer:

The product of mass and velocity

Explanation:

I  just did it and got it right with a 100%

Posted 1/3/23

From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.

Answers

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 4.35 m above where it started. Using conservation of energy, find the height of the ball when it has a speed of 2.5 m/s.

Answers

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m and we assume fully developed internal flow, find the pressure drop across this pipe length.

Answers

Answer:

[tex]\triangle P=1.95*10^{-4}[/tex]

Explanation:

Mass [tex]m=0.001[/tex]

Diameter [tex]d=1.2m[/tex]

Length [tex]l=10m[/tex]

Generally the equation for Volume flow rate is mathematically given by

 [tex]Q=AV[/tex]

 [tex]V=\frac{Q}{\pi/4D^2}[/tex]

 [tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]

 [tex]V=8.84*10^{-4}[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]F=\frac{64}{Re}[/tex]

Where Re

Re=Reynolds Number

 [tex]Re=\frac{pVD}{\mu}[/tex]

 [tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]

 [tex]Re=1040[/tex]

Therefore

 [tex]F=\frac{64}{Re}[/tex]

 [tex]F=\frac{64}{1040}[/tex]

 [tex]F=0.06[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]Head loss=\frac{fLv^2}{2dg}[/tex]

 [tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]

 [tex]H=19.9*10^{-9}[/tex]

Where

[tex]H=\frac{\triangle P}{\rho g}[/tex]

[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]

[tex]\triangle P=H*\rho g[/tex]

[tex]\triangle P=1.95*10^{-4}[/tex]

 

how can the starch be removed from the leaves of potted plants​

Answers

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

what are the symptoms of hepatitis 'b'​

Answers

Fever might be one of the signs of acute hepatitis B.
Fatigue is a common occurrence.
Appetite loss is common.
Nausea is a feeling of nausea.
Pain in the abdomen.
Urine that is dark in color.
Bowel motions that are clay-colored.

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting

Answers

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong

Answers

A. Because the third laws say that for every action force the is an equal and opposite reaction force

[tex] \huge \mathfrak{Answer.... }[/tex]

The Correct Answer is :

B. if there is even a slight imbalance in third law there will be a net force causing acceleration.

A slight difference in the forces can result in acceleration of an object.

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The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.​

Answers

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

what is the application of a spherometer in the medical field?​

Answers

Answer:

To correct the defects of vision by measuring the radius of curvature and thus the power of the lenses.

Explanation:

A spherometer is an instrument used to measure the curvature of objects such as lenses and curved mirrors.

Generally it consists of a fine screw which is moving in a nut carried on the center of a 3 small legged table or frame. The feet forms the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres.

If the screw has two turns of the thread to the milli meter the head is generally divided into 50 equal parts, so that differences of 0.01 millimeter may be measured without using a vernier scale.

The spherometer is used to measure the radius of curvature of the lenses so that the opthalmologist find the focal length of the lens and then give the  power to the lens to correct the defects of vision.  

define emperical formula and what is the dimensional formula of force and energy​

Answers

Answer:

An empirical formula represents the simplest whole number ratio of various atoms present in a compound.The dimensional formula of force is [[tex]MLT^{-2}[/tex]]The dimensional formula of energy is [[tex]ML^{2} T^{-2}[/tex]]

Calculate the battery voltage to send 2.5 A of current through a light bulb with 3.6 ohms of resistance

Answers

Answer:

9 volts

Explanation:

[tex]v = i \times r \\ v = 2 .5 \times 3.6 \\ v = 9[/tex]

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